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Proof step 22 from proof valid-global-sqn-next-max (previous step)

apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), m < # apebs, m1 < n, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), n ≤ valid-sqns(v0, m5, apebs, m).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) = ∅, v0 = v ∧ m5 = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
apebs[m] = mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false), m1 < n, m < # apebs, m1 < valid-sqns(v0, m5, apebs, m).max, valid-peb(mkpeb(aehdr, avidhdr(v, m0, m1, m2, m4), buf, m3, false)), n ≤ valid-sqns(v0, m5, apebs, m).max, ∀ v0, m5. valid-sqns(v0, m5, apebs, m) ≠ ∅ → valid-sqns(v0, m5, apebs, m).max < n ⊦ valid-sqns(v0, m5, apebs, m) = ∅, v0 = v ∧ m5 = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic